/*
All square roots are periodic when written as continued fractions and can be written in the form:

√N = a0 +
1

 
a1 +
1

 
 
a2 +
1

 
 
 
a3 + ...


For example, let us consider √23:

√23 = 4 + √23 — 4 = 4 + 
1
 = 4 + 
1

 
1√23—4
 
1 + 
√23 – 37


If we continue we would get the following expansion:

√23 = 4 +
1

 
1 +
1

 
 
3 +
1

 
 
 
1 +
1

 
 
 
 
8 + ...


The process can be summarised as follows:

a0 = 4,
 
1√23—4
 = 
√23+47
 = 1 + 
√23—37

a1 = 1,
 
7√23—3
 = 
7(√23+3)14
 = 3 + 
√23—32

a2 = 3,
 
2√23—3
 = 
2(√23+3)14
 = 1 + 
√23—47

a3 = 1,
 
7√23—4
 = 
7(√23+4)7
 = 8 + 
√23—4

a4 = 8,
 
1√23—4
 = 
√23+47
 = 1 + 
√23—37

a5 = 1,
 
7√23—3
 = 
7(√23+3)14
 = 3 + 
√23—32

a6 = 3,
 
2√23—3
 = 
2(√23+3)14
 = 1 + 
√23—47

a7 = 1,
 
7√23—4
 = 
7(√23+4)7
 = 8 + 
√23—4


It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:
√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5
Exactly four continued fractions, for N ≤ 13, have an odd period.
How many continued fractions for N ≤ 10000 have an odd period?

Anser:
Time:
*/
package main

import (
	"fmt"
	"time"
)

func main() {
	tstart := time.Now()



	tend := time.Now()
	fmt.Println(tend.Sub(tstart))
}